 Solution

Write $U^{\ast }={\overline{U}}^T$ for the conjugate transpose. Taking determinants in $U^{\ast }AU=A$ gives $|\det U{|}^2\det A=\det A$. Since the Hermitian matrix $A$ is invertible, every such $U$ is invertible and $|\det U|=1$. The identity belongs to $U_A$; if $U,V\in U_A$, then $(UV{)}^{\ast }A(UV)=V^{\ast }AV=A$; and multiplying $U^{\ast }AU=A$ by $(U^{\ast }{)}^{-1}$ and $U^{-1}$ gives $(U^{-1}{)}^{\ast }A{U}^{-1}=A$. Associativity comes from matrix multiplication, so this is a group.

The Hermitian form $h_A(v,w)=v^{\ast }Aw$ obeys $h_A(Uv,Uw)=h_A(v,w)$ precisely when $U\in U_A$. Thus $U_A$ is the group of linear isometries of this nondegenerate Hermitian form; positivity of $A$ is unnecessary.

For $A=I$, $U$ is a unitary matrix. Here is a direct diagonalization of a matrix proof. Over $\mathbb{C}$, its characteristic polynomial has a root, giving a unit eigenvector $v$ with $Uv=\lambda v$. Preservation of the inner product gives $|\lambda |=1$. If $w\perp v$, then $⟨Uw,Uv⟩=⟨w,v⟩=0$, so $v^{\perp }$ is invariant. The restriction is again a unitary matrix. Induction on the dimension supplies an orthonormal basis of eigenvectors. In that basis, $U$ is diagonal, with diagonal entries on the unit circle.